Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int sy, a, b;
int n, q;
int l[200010];
int r[200010];
int sum[200100];
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  cin >> sy >> a >> b;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    if (i) {
      sum[i] = sum[i - 1] + l[i] - r[i - 1];
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x1, y1;
    cin >> x1 >> y1;
    int x2 = a, y2 = sy;
    int A = y2 - y1;
    long long C = 1ll * x2 * y1 - 1ll * x1 * y2;
    A *= -1, C *= -1;
    int lo = 0, hi = n;
    while (lo != hi) {
      int mi = (lo + hi) >> 1;
      if (-C <= 1ll * r[mi] * A)
        hi = mi;
      else
        lo = mi + 1;
    }
    int L = lo;
    x2 = b, y2 = sy;
    A = y2 - y1, C = 1ll * x2 * y1 - 1ll * x1 * y2;
    A *= -1, C *= -1;
    lo = -1, hi = n - 1;
    while (lo != hi) {
      int mi = (lo + hi + 1) >> 1;
      if (-C >= 1ll * l[mi] * A)
        lo = mi;
      else
        hi = mi - 1;
    }
    int R = hi;
    if (R == -1) {
      cout << 0.0 << '\n';
    } else if (L == n) {
      cout << 0.0 << '\n';
    } else {
      double mul = 1.0 * abs(sy - y1) / y1;
      double an = sum[R] - sum[L];
      an *= mul;
      x2 = a, y2 = sy;
      A = y2 - y1, C = 1ll * x2 * y1 - 1ll * x1 * y2;
      double x3 = -1.0 * C / A;
      if (x3 < l[L]) an += (1.0 * l[L] - x3) * mul;
      x2 = b, y2 = sy;
      A = y2 - y1, C = 1ll * x2 * y1 - 1ll * x1 * y2;
      x3 = -1.0 * C / A;
      if (r[R] < x3) an += (x3 - r[R]) * mul;
      cout << 1.0 * (b - a) - an << '\n';
    }
  }
}